THE YAMABE EQUATION ON COMPLETE MANIFOLDS WITH FINITE VOLUME

arXiv:1111.2471v1 [math.DG] 10 Nov 2011

FINITE VOLUME

NADINE GROSSE

Abstract. We prove the existence of a solution of the Yamabe equation on complete manifolds with finite volume and positive Yamabe invariant. In order to circumvent the standard methods on closed manifolds which heavily rely on global (compact) Sobolev embeddings we approximate the solution by eigenfunctions of certain conformal complete metrics.

This also gives rise to a new proof of the well-known result for closed manifolds and positive Yamabe invariant.

1. Introduction

Yamabe examined whether a closed n-dimensional Riemannian manifold (M, g) (n ≥ 3) possesses a metricgconformal togwith constant scalar curvature. His striking idea was the consideration of the so-called Yamabe invariant, see Definition 1, which gave the possibility to view the question as a variational problem. Works from Aubin [1], Schoen [10] and Trudinger [14] answered the question of Yamabe affirmatively.

There are several possibilities to generalize the Yamabe problem to open (i.e. noncompact and without boundary) manifolds.

One possibility is simply to pose the same question as Yamabe did. On open manifolds, this gives much more freedom. We want to make this more precise by comparing to the closed case:

On closed manifolds, if the Yamabe invariantQis nonpositive, then every conformal metric having constant scalar curvaturec and unit volume fulfillsc =Q. In case thatQ <0 this conformal metric is even unique. If Q > 0 and there is a conformal metric with constant scalar curvaturec, one see immediately that c≥Q. But nevertheless, on closed manifolds in all cases a conformal metric with constant scalar curvature has the same sign as the corresponding Yamabe invariant.

On open manifolds, this is no longer true, an easy example is given by an open ball in the Euclidean space. Its Yamabe invariant is the one of the standard sphere, but it carries conformal metrics of constant scalar curvature of all signs: The original Euclidean metric has zero scalar curvature, the spherical metric has constant positive and the hyperbolic metric has constant negative scalar curvature. But all those metrics are conformally equivalent.

That’s why the question is often posed more restrictively. A first possibility is to fix the sign of the constant scalar curvature and/or ask additionally for completeness. This was done by many authors and many results with positive and negative answers were obtained, see for example [2], [7], [15].

2000Mathematics Subject Classification. 53C21, 53A30, 35R01.

Key words and phrases. Yamabe problem, scalar curvature, complete manifolds of finite volume.

I like to thank the Hausdorffcenter in Bonn where parts of this paper were written down for its hospitality.

1

The second is to stick to the original Yamabe problem and ask for solutions to the Euler- Lagrange equation of the Yamabe problem, i.e. unit volume metrics with constant scalar curvature Q. This version was studied for example for manifolds with bounded geometry and positive scalar curvature in [8] using a compact exhaustion of the open manifold and for manifolds bounded geometry and positive Yamabe invariant in [5] using weighted Sobolev embeddings.

In this paper we consider the second type of a noncompact Yamabe problem on complete manifolds of finite volume.

Let (Mⁿ, g) be ann-dimensional complete connected Riemannian manifold of finite volume and n ≥3. Let Lg =an∆g+ scalg be the conformal Laplacian where scalg is the scalar curvature of the metric gandan = 4ⁿ⁻¹_n−2.

Definition 1. The Yamabe invariant of(M, g)is given by

Q(M, g) = inf (

Qg(v) :=

R

MvLgvdvolg

kvk²_Lp(g)

v∈C_c^∞(M), v6= 0 )

where p= _n−2²ⁿ andC_c^∞(M)denotes the set of compactly supported real valued functions on M.

Q is conformally invariant which is seen from the conformal transformation formula of the conformal Laplacian: For g=f²g wheref ∈C_>0^∞(M) is a smooth positive real function on M we have

Lgv=f Lgv wherev=f⁻ⁿ⁻²²v.

The Yamabe invariant is given as a variational problem. Its Euler-Lagrange equation is Lgv=Qv^p−1 v∈H₁², kvkL^p(g)= 1. (1) The aim of this paper is to study the existence of a smooth positive solution of (1) for complete manifolds of finite volume.

The standard proof for the Yamabe problem on closed manifolds heavily relies on the exis- tence of compact Sobolev embeddings. On complete open manifolds of finite volume there do not even exist continuous Sobolev embeddings H₁²֒→L^p, [6, Lem. 3.2]. That’s why we will use a different approach by approximating the desired solution by certain eigenfunctions of conformal metrics, cp. Section 3.

In the standard proof on closed manifolds one uses the subcritical Yamabe problem to get solutions of differential equations that are somehow ’near’ to the desired Euler-Lagrange equations. This allows to show converges of a sequence of those solutions which then serve as test functions for the critical problem. In our approach here the eigenfunctions will play the role of these subcritical solutions and we obtain

Theorem 2. Let(M, g)be an open complete manifold of finite volume with 0< Q(M, g)<

Q(M, g)andk(scalg)−kⁿ₂ <∞where (scalg)−:=−min{scal,0}.

Then, there exists a smooth positive solution v∈H₁² of Lgv=Qv^p−1 withkvkL^p(g)= 1.

Q(M, g) is the Yamabe invariant at infinity, see Definition 4, and replacesQ(Sⁿ) that appears at this point in the closed case, cf. Remark 5.

The non-existence of a continuous Sobolev embeddingH₁²֒→L^phas the following straight- forward implications: If Q > 0, scalg cannot be bounded from above. Moreover, if v is a solution as in Theorem 2, g =vⁿ⁻²ⁿ²g is a metric with finite volume and constant scalar curvature and for all v ∈ C_c^∞(M) kvkL^p(g) ≤(max{an, Q})¹²kvkH²₁(g). Thus,g cannot be complete.

The method used to prove Theorem 2 also gives rise to a different proof for the closed case with positive Yamabe invariant, see Theorem 14. Moreover, the method can be adapted to similar contexts, e.g one can obtain similar results for the spinorial Yamabe invariant, cf.

[4].

2. Preliminaries In this section we collect some facts on the Yamabe invariant.

Remark 3. On complete manifolds instead of taking the infimum overC_c^∞in the definition of the Yamabe invariant 1 one could as well take the infimum overv ∈ L²∩H_1,loc² with

R

MvLgvdvolg

<∞. This is seen when consideringQg(ηiv) for suitable cut-off functions ηi withηi→1.

Definition 4. [9]Let(M, g)be an openn-dimensional manifold with a compact exhaustion Ki fulfilling Ki ⊂ Ki+1 M and ∪iKi = M. Then the Yamabe invariant at infinity is defined as

Q(M, g) := lim

i→∞Q(M\Ki, g).

Note that Q(M \Ki, g) ≤ Q(M \Ki+1, g) since when considering only a subset less test functions can be used in Definition 1. Together withQ(M, g)≤Q(Sⁿ) [11] whereQ(Sⁿ) = n(n−1)vol(Sⁿ)²ⁿis the Yamabe invariant of the sphere with the standard metric the sequence Q(M \Ki, g) is monotonically increasing and bounded. Thus,Qalways exists and it holds Q(M, g)≤Q(Sⁿ). Furthermore,Qdoes not depend on the choice of the sequenceKi. Remark 5. We note that the conditionQ(M, g)< Q(Sⁿ) in Theorem 2 replacesQ(M, g)<

Q(Sⁿ) that appears in the closed case. This can be seen since forp∈M we haveQ(M, g) = Q(M\ {p}, g) [11, Lem. 2.1] andQ(M\ {p}, g) = limi→∞Q(Bǫ(p), g) =Q(Sⁿ) whereBǫ(p) is a ball aroundpwith radiusǫ.

The blow-up argument in the standard proof of the Yamabe problem [12] which rules out concentration phenomena at a fixed point shows that for fixed x ∈ M Q(Bǫ(x), g) → Q(Rⁿ, gE) =Q(Sⁿ, gst) asǫ→0. We will need the following slight generalization:

Lemma 6. For all compact subsets U ⊂M andδ >0 there is anǫ=ǫ(U, δ)>0 such that for allx∈U: Q(Bǫ(x), g)≥Q(Sⁿ)−δ.

Proof. LetU andδbe fixed. Then for eachx∈U letǫ(x) be the maximal radius such that Q(Bǫ(x), g)≥Q(Sⁿ)−δ is fulfilled. Setǫ= infX∈Uǫ(x). Suppose ǫ= 0. Then there is a sequence xi ∈ U withǫ(xi) →0. Since U is compact, xi → x ∈U. Note that on closed manifolds Q depends smoothly on g in the C²-topology [3, Proof of Prop. 7.2.]. Thus, ǫ(xi)→ǫ(x)>0 which is a contradiction. Thus,ǫ >0.

3. Nonnegative Yamabe invariants and the L²-spectrum On closed manifolds and ifQ≥0,

Q(M, g) = inf{µ(Lg)|g∈[g]}

whereµ(Lg) is the lowest eigenvalue of the conformal LaplacianLgand [g] :={g=f²g|f ∈ C_>0^∞(M)} denotes the conformal class ofg.

On general manifolds the spectrum ofLg does not only contain eigenvalues but there can be residual and continuous spectrum. Moreover, in generalLg is even not essentially self- adjoint.

We consider

µ(Lg) = inf ( R

MvLgvdvolg

kvk²_L2(g)

v∈C_c^∞(M) )

.

IfLg is essentially self-adjoint,µ(Lg) is the minimum of the spectrum ofLg. Remark 7. IfQ≥0 and vol(M, g) = 1, thenR

MvLgvdvolg ≥Qkvk²_p≥Qkvk²₂, i.e. µ(Lg)≥ QandLgis bounded from below. Then,Lgis essentially self-adjoint onC_c^∞(M), [13, Thm.

1.1] and possesses only eigenvalues and essential spectrum. Moreover, the spectrum is real.

If it is clear from the context to which Riemannian manifold (M, g) we refer, we abbreviate k.ks:=k.kL^s(g).

Lemma 8. Let (M, g)be a Riemannian manifold withQ≥0. Then Q(M, g) = inf{µ(Lg)| g∈[g],vol(M, g) = 1}.

If (M, g)has additionally unit volume,

Q(M, g) = inf{µ(Lg) |g=f²g,vol(M, g) = 1, ∃ a compact subset Kf⊂M : f|M\Kf = 1}.

If for a function f ∈C_>0^∞(M) such a compact subsetKf exists, we shortly say that f ≡1 near infinity. The proof of the first part is the same as in the closed case. But since we are not aware of a reference we shortly give the proof.

Proof. Without loss of generality we can assume that g already has unit volume. Since Q≥0,R

MvLgvdvolg≥0 for allv∈C_c^∞(M).

From Remark 7 we haveµ(Lg)≥Q(M, g) for all conformal metricsg∈[g] with unit volume.

On the other hand, let vi ∈ C_c^∞(M) be a minimizing sequence for Q with kvikp = 1 and R

MviLgvidvolg → Q. Set gi = kvi+i⁻¹k⁻¹_p (vi+i⁻¹)_n−2⁴

g. Then, vol(M, gi) = R

M kvi+i⁻¹k⁻¹_p (vi+i⁻¹)^p

dvolg = 1 (Note that kvi+i⁻¹kp ≤ kvikp+i⁻¹ = 1 +i⁻¹ is finite.). Moreover,

kvik²_L2(gi)= Z

M

kvi+i⁻¹k⁻¹_p (vi+i⁻¹)_n−⁴2

v_i²dvolg

≥ Z

M

(1 +i⁻¹)⁻ⁿ⁻²⁴ (vi+i⁻¹)ⁿ⁻²⁴ v²_idvolg

≥(1 +i⁻¹)⁻ⁿ⁻²⁴ Z

M

v_i^pdvolg= (1 +i⁻¹)⁻ⁿ⁻²⁴ . Hence,

0≤µ(Lg_i)≤ R

MviLg_ividvolg_i

kvik²_L2(gi)

≤ R

MviLgvidvolg

(1 +i⁻¹)⁻ⁿ⁻⁴² →Q(M, g) as i→ ∞which finishes the proof of the first claim.

Let now (M, g) be complete and of finite volume andvibe the test sequence of above. LetKi

be a sequence of compact subsets with supp vi⊂Ki. The valuekvik⁻²_L2(gi)

R

MviLg_ividvolg_i

only depends on the metricgi on suppvi. Thus, we can deformgisuch that the conformal factor fi = 1 outside a compact subset Kf_i with Ki ⊂⊂ Kf_i M, f_i²g ≡ gi on Ki and vol(fi²g) = 1.

In particular, if g was complete, all thoseg are also complete.

Next we study the Yamabe invariant if essential spectrum is present.

Lemma 9. Let (M, g) be a complete Riemannian manifold of unit volume. Let Lg be essentially self-adjoint onC_c^∞(M)and let the essential spectrum ofLgbe non-empty. Then Q(M, g) =Q(M, g)≤0.

Proof. Letµbe in the essential spectrum ofLg. Then there is a sequencevi ∈C_c^∞(M\Bi) where Bi a ball with radius i around a fixed point z ∈ M such that k(Lg−µ)vik2 → 0, kvik2= 1 andvi→0 weakly in L². Then, using 1 =kvik2≤ kvikpvol(M \Bi)ⁿ² and, thus, kvikp≥1 we estimate

Q(M \Bi)≤ R

MviLgvidvolg

kvik²_p ≤kLgvik2kvik2

kvik²_p ≤ (k(Lg−µ)vik2+|µ|)kvik2

kvik²_p

≤ (k(Lg−µ)vik2+|µ|)kvikpvol(M\Bi)ⁿ² kvik²p

≤(k(Lg−µ)vik2+|µ|)vol(M\Bi)ⁿ²

where the right hand-side goes to zero asi→ ∞.

From that and Remark 7 it follows directly

Corollary 10. If(M, g)is a complete Riemannian manifold of unit volume withQ(M, g)>

0 or Q(M, g) > Q(M, g) = 0, there exists a sequence of gi = f_i²g with fi ∈ C_>0^∞(M), R

Mf_iⁿdvolg= 1 and eigenvaluesµi:=µ(Lg_i)→Q asi→ ∞.

4. Proof of Theorem 2

From Corollary 10 we have: If Q > 0, there exists a sequence of eigenfunctions vi with Lgivi = µivi and R

M|vi|²dvolgi = 1. vi is eigenfunction to the lowest eigenvalue µi of gi = f_i²g (fi = 1 near infinity) and, hence, positive. Viewing these equation w.r.t. the reference metricg we obtain the following setting

Lgvi=µif_i²vi

Z

M

f_i²v²_idvolg = 1, Z

M

f_iⁿdvolg= 1, µiցQ, vi>0.

Firstly we note thatR

M|vi|²dvolg_i =R

Mf_i²v_i²dvolg_i= 1 andfi= 1 outside a compact subset impliesvi∈L²(g).

Moreover, due to Remark 3 vi can serve as a test function for Q and, thus, Qkvik²_p ≤ R

MviLgvidvolg = µi. Since Q > 0 vi ∈L^p(g) and if then i → ∞, we obtain kvikp →1.

Thus,vi is uniformly bounded inL^p(g) and, due to the finite volume, also inL²(g).

From

µi= Z

M

viLgvidvolg =ankdvik²₂+ Z

M

scalgv²_idvolg

≥ankdvik²₂− Z

M

(scalg)−v_i²dvolg≥ankdvik²₂− k(scalg)−kⁿ₂kvik²_p

and the assumption thatk(scalg)−kⁿ₂ <∞ we see thatkdvik2 is also uniformly bounded.

Summarizingvi is uniformly bounded in H₁² and, hence,vi →v ≥0 weakly in H₁² and in L^p. Moreover,R

Mf_iⁿdvolg = 1 implies that there is f ∈ Lⁿ such thatf_i² →f² weakly in Lⁿ².

Lemma 11. Let f_i²→f² weakly inLⁿ² and vi→v weakly inH₁².

i) Then f_i²vi→f²v inL^s(U)for all compact subsetsU ⊂M and1< s < q= _n+2²ⁿ . ii) If additionally Lgvi =µif_i²vi andµi→Q,f andv weakly fulfill Lgv=Qf²v.

Proof. i) We fixw∈L^s∗(U) with ¹_s+_s¹∗ = 1. Then

Z

U

(fi²vi−f²v)wdvolg

≤ Z

U

|fi²−f²| |vw|dvolg+ Z

U

f_i²|vi−v| |w|dvolg

The weak convergence vi →v in H₁² implies strong convergence vi → v on L^p′(U) for all 1≤p^′ < p. We chooseq^′ such thatp > q^′> _n−2ⁿ . Then H¨older inequality implies

kvwk_Ln−ⁿ_2(U)≤ kvk_L^q′_(U)kwk

L

nq′

(n−2)q′ −n(U)

<∞

if _(n−2)q^nq′′−n ≤s^∗. The choice ofq^′ impliesp < _(n−2)q^nq′_′−n <∞ and, thus,p < s^∗<∞ and 1< s < q.

Hence,R

U|f_i²−f²| |vw|dvolg →0 asi→ ∞. For the second summand of the above inequality we have

Z

U

f_i²|vi−v| |w|dvolg≤ kvi−vk_L^q′_(U)kfi²wk

L

q′

q′ −1(U)

≤ kvi−vk_L^q′_(U)kfik²_Ln(U)kwk

L

nq′

(n−2)q′ −n(U)

≤ kvi−vk_L^q′_(U)kwkL^s(U)

→0 asi→ ∞.

ii) Letw∈Cc^∞(M).

Z

M

(Lgv−Qf²v)wdvolg

= Z

M

(Lgv−Lgvi+µif_i²vi−Qf²v)wdvolg

≤an

Z

M

(dv−dvi)dwdvolg

+ Z

M

(v−vi)(scalgw)dvolg

+µi

Z

M

(f_i²vi−f²v)wdvolg

+|Q−µi| Z

M

f²|vw|dvolg

All summands on the right-hand side tend to zero as i → ∞ sincevi → v weakly in H₁² (note that scalgw∈C_c^∞⊂L²), part i) andµi →Q.

In order to finish the proof of Theorem 2 it remains to show that f²=v^p−2 andkvkp= 1.

We start with a non-vanishing result.

Lemma 12. In the setting of Lemma 11 and assuming 0< Q(M, g)< Q(Sⁿ),v does not vanish identically.

Proof. We prove by contradiction and assume v ≡ 0 and, hence, R

Uv^s_idvolg → 0 for all compact subsetsU ⊂M and 1≤s < p.

Firstly, we want to show that then alsoR

Uv_i^pdvolg→0 for all compact subsetsU ⊂M. For that, we assume the contrary, i.e. kvikL^p(U) > C(U)>0 and consider small balls B2ǫ(x) with x∈M. We chooseǫsmall enough such that for allx∈U Q(B2ǫ(x))> Q(M, g). Due to Q(Sⁿ) > Q(M, g) and Lemma 6 this is always possible. Then we cover U by finitely many of those ballsB2ǫ(x) and define smooth cut-off functionsηǫ,x compactly supported in B2ǫ(x) that are 1 onBǫ(x) and|dηǫ,x| ≤2ǫ⁻¹. Then we estimate

Q(B2ǫ(x), g)≤ R

B_2ǫηǫ,xviLg(ηǫ,xvi)dvolg

R

B2ǫ(x)η^pǫ,xv^p_idvolg

_p² = R

B_2ǫη_ǫ,x² viLgvidvolg+anR

B_2ǫ|dηǫ,x|²v²_idvolg

R

B2ǫ(x)η^pǫ,xv_i^pdvolg

²_p

≤µi

R

B2ǫ(x)η_ǫ,x² f_i²v²_idvolg+an4 ǫ²

R

B2ǫ(x)v²_i R

B_2ǫ(x)ηǫ,x^p v^p_idvolg

²_p ≤µi+an

4 ǫ²

R

B2ǫ(x)v²_idvolg

R

B_ǫ(x)v_i^pdvolg

²_p

≤µi+an

4 ǫ²C(U)²

Z

B2ǫ(x)

v²_idvolg

where in the second last step we used the H¨older inequality to estimate the summand including µi. If i tends to ∞, we obtain Q(B2ǫ(x), g) ≤ Q which is a contradiction to Q(B2ǫ(x))> Q. Thus,kvikL^p(U)→0 asi→ ∞.

Next, letχR be a smooth cut-off function withχR= 0 on BR:=BR(z) for a fixedz∈M, χR= 1 onM \B2R and|dχR| ≤2R⁻¹. Then

Q= lim

i→∞

Z

M

viLgvidvolg= lim

i→∞

Z

M

χ²_RviLgvidvolg+µi

Z

M

(1−χ²_R)f_i²v²_idvolg

≥ lim

i→∞

Z

M

χ²_RviLgvidvolg+µi

Z

BR

f_i²v_i²dvolg

≥ lim

i→∞

Z

M

χRviLg(χRvi)dvolg−an

Z

M

|dχR|²v_i²dvolg+µi

Z

BR

f_i²v_i²dvolg

≥ lim

i→∞

Q(M \BR, g)kχRvik²p−4an

R²kvik²_L2(B2R)+µi

Z

B_R

f_i²v²_idvolg

≥ lim

i→∞

Q(M \BR, g)(kvikp− k(1−χR)vikp)²−4an

R²kvik²_L²_(B2R)+µi

Z

B_R

f_i²v²_idvolg

≥ lim

i→∞

Q(M \BR, g) kvikp− kvikL^p(B2R)2

−4an

R²kvik²_L2(B2R)+µi

Z

B_R

f_i²v_i²dvolg

WithkvikL^s(U)→0 for 1≤s≤pon compact subsets U⊂M,kvikp→1 and Z

B_R

f_i²v_i²dvolg≤ kfiknkvik²_Lp(B_R)≤ kvik²_Lp(B_R)→0 we obtain for allRthat

Q(M, g)≥Q(M\BR, g).

That contradictsQ > Q. Thus,v6= 0.

Now we can estimate

Q≤ R

MvLgvdvolg

kvk²p

≤Q R

Mf²v²dvolg

kvk²p

≤Qkfk²_nkvk²_p kvk²p

≤Q.

Hence, there is already equality. In particular, from the equality case in the used H¨older inequality we getf²=v^p−2and 1 =kfkn=kvkp. Smoothness ofvis obtained by standard local elliptic regularity theory. By the maximum principle one sees that v is everywhere

positive which concludes the proof of Theorem 2.

Standard local elliptic regularity also gives thatv is locally inC^2,α.

Remark 13 (On the assumption on the scalar curvature).

In Theorem 2 we assume thatk(scalg)−k_Lⁿ_2(g)<∞. If the Yamabe invariantQ(g) =−∞, this could never be true. But in general it can happen that even thoughk(scalg)−k_Lⁿ_2(g)=

∞, Q is finite and even positive. The easiest example is the standard hyperbolic space Hⁿ which has constant negative scalar curvature, infinite volume but the Yamabe invariant of the standard sphere. From this we can even easily construct an example with finite volume: Firstly, we note thatk(scalg)−k_Lⁿ_2(g)is scale invariant. Let us take a ballB in the hyperbolic space with k(scalg)−k_Lⁿ_2(g) = 1 and then rescale it such that the rescaled ball Bi has volume i⁻². If we consider the disjoint sum of the Bi, we obtain an example for a (disconnected) Riemannian manifold of finite volume andk(scalg)−k_Lⁿ_2(g)=∞.

We assume that Q(g) = −∞ if and only if k(scalg)−k_Lⁿ_2(g) = ∞ for all g ∈ [g]. But unfortunately we still cannot prove this. Even if this is true, this alone does not help in our context since we need a complete metric of finite volume with k(scalg)−k_Lⁿ_2(g)<∞which probably cannot be achieved in general.

5. On closed manifolds

The method we used in Theorem 2 for complete manifolds of finite volume allows to reprove the result on closed manifolds with positive Yamabe invariant.

Theorem 14. Let (M, g) be a closed n-dimensional Riemannian manifold with 0 < Q <

Q(Sⁿ). Then, there is a smooth positive solutionv ∈H₁² of the Yamabe equation (1).

Proof. The proof in the closed case is essentially the same as the one presented in Section 4.

The only little difference occurs in the proof of Lemma 12 where the cut-off function χR is introduced andQis estimated. We make the following change – we take the smooth cut-off function ηǫ introduced before in Lemma 12. Then with the same estimate as in Lemma 12, where M\Bǫ substitutesB2R andM\B2ǫreplacesBR, we obtain

Q≥ lim

i→∞ Q(B2ǫ, g)(kvikp− kvikL^p(M\B_ǫ))²−4an

ǫ² kvikL²(M\B_ǫ)+µi

Z

M\B2ǫ

f_i²v²_idvolg

!

=Q(B2ǫ, g)

For ǫ small enough this gives a contradiction to Q(M)< Q(Sⁿ) due to Lemma 6. Thus, following the rest of the proof in Section 4 we obtain thatv is a smooth positive solution of Lgv=Qv^p−1withkvkp= 1. Note that on closed manifolds the conditionk(scalg)−k_Lⁿ_2(g)<

∞of Theorem 2 is trivially fulfilled.

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